How to solve $\sum_{i=1}^{n} \sin(x_i - \mu) = 0$ for $\mu$? (Maximum likelihood estimation)

Question

The distribution is defined as:

$f(x; k, \mu) = exp(k\cos(x-\mu))$ (I have removed the term before it because it will be $0$ when we take the derivative of the log). We must find the MLE of $\mu$

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My work so far:

• $\prod\limits_{i=1}^n exp(k\cos(x_i - \mu)) = exp(k\cos(x_1 - \mu)k\cos(x_2 - \mu)...k\cos(x_n - \mu)) = \exp(k\sum_{i=1}^{n} \cos(x_i - \mu))$

• taking log: $=k\sum_{i=1}^{n} \cos(x_i - \mu)$

• taking derivative wrt $\mu$: $=k\sum_{i=1}^{n} \sin(x_i - \mu)$

• setting to $0$: --> $\sum_{i=1}^{n} \sin(x_i - \mu) = 0$

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I have no idea how to proceed from here. Is there anything I am missing or any calculation errors?

Answer 1

I guess using the last point of your question, you can use the definition of $\sin(A - B) = \sin(A) \cos (B) - \cos(A) \sin(B)$. Thus, $$ \sum_i \sin(x_i - \mu) = \sum_i [\sin(x_i) \cos(\mu) - \cos(x_i)\sin(\mu)] = 0$$ now, after some trivial algebra you should get (if and only if $\mu \neq 0$ and $\sum_i \cos(x_i) \neq 0$) $$\hat{\mu} = \tan^{-1}\left( \frac{\sum_{i=1}^n \sin (x_i)}{\sum_{i=1}^n \cos (x_i)} \right)$$ which you can now easily compute for any $x_i$'s.